Question

Prove that the eigenvalues of an upper triangular matrix T are the entries of its main diagonal, so that the trace of T is the sum of its eigenvalues.

Answer 1

Eigenvalues are roots ofpolynomial det(T-kI)=0, where I is an identity matrix.
Since T is upper triangular matrix, then we deal with determinant of upper triangular
matrix again, and its determinant is product of diagonal entries.
So, k is one of the roots of (a_11-k)(a_22-k)...(a_nn-k) = 0.
Then all k varies over set {a_11,...,a_nn}, and sum of all k's is
a_11+....+a_nn=tr(T).